Argon is an inert gas used in light bulbs to retard the vaporization of the filament. A certain light-bulb containing argon at 1.25 atm and `18^(@)C` is heated to `85^(@)C` at constant volume. Calculate its final pressure.

Options

A. 1.53 atm
B. 1.25 atm
C. 1.35 atm
D. 2 atm

Answer

To solve the problem of calculating the final pressure of argon gas in a light bulb when heated from 18°C to 85°C at constant volume, we can use the relationship between pressure and temperature given by Gay-Lussac's Law. Here’s the step-by-step solution:

### Step 1: Understand the relationship
Since the volume is constant and the number of moles of gas does not change, we can use the formula:
\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \]
where \( P_1 \) is the initial pressure, \( P_2 \) is the final pressure, \( T_1 \) is the initial temperature in Kelvin, and \( T_2 \) is the final temperature in Kelvin.

### Step 2: Convert temperatures from Celsius to Kelvin
To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature.

- Initial temperature \( T_1 \):
\[ T_1 = 18 + 273.15 = 291.15 \, K \approx 291 \, K \]

- Final temperature \( T_2 \):
\[ T_2 = 85 + 273.15 = 358.15 \, K \approx 358 \, K \]

### Step 3: Substitute known values into the equation
We know:
- \( P_1 = 1.25 \, \text{atm} \)
- \( T_1 = 291 \, K \)
- \( T_2 = 358 \, K \)

Now, substituting these values into the equation:
\[ \frac{P_2}{1.25} = \frac{358}{291} \]

### Step 4: Solve for \( P_2 \)
To find \( P_2 \), we rearrange the equation:
\[ P_2 = 1.25 \times \frac{358}{291} \]

Calculating the right-hand side:
\[ P_2 = 1.25 \times 1.229 = 1.536 \, \text{atm} \approx 1.54 \, \text{atm} \]

### Final Answer
The final pressure \( P_2 \) is approximately **1.54 atm**.

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